JEE Physics Electrostatic Potential and Capacitance Notes Set 05

Capacitance of an Isolated Conductor

A conductor of finite size raised to potential V by a charge q satisfies \( q=CV \); the constant C is its capacitance. It is a scalar, measured in farad (1 F = 1 C/V), with dimensions \( M^{-1}L^{-2}T^{4}A^{2} \).
C depends on the size and shape of the conductor, the surrounding medium (it rises by the factor K in a dielectric) and nearby conductors. It does not depend on q or V.
An isolated sphere of radius R has \( C=4\pi\epsilon_0R \), so the earth (R = 6400 km) has a capacitance of only about 711 \( \mu \)F.
Energy stored: \( U=\frac{q^2}{2C}=\frac12CV^2=\frac12qV \), spread through space with energy density \( \frac12\epsilon_0E^2 \) (or \( \frac12\epsilon_0\epsilon_rE^2 \) in a medium).

Sharing of Charge Between Conductors

When two charged conductors are joined by a wire, charge moves until the potentials are equal. Total charge is conserved:
\( V=\frac{Q_1+Q_2}{C_1+C_2},\quad Q_1'=\frac{C_1}{C_1+C_2}(Q_1+Q_2),\quad Q_2'=\frac{C_2}{C_1+C_2}(Q_1+Q_2) \).
Energy lost as heat: \( H=\frac12\frac{C_1C_2}{C_1+C_2}(V_1-V_2)^2 \). Always put the signs of the charges and potentials in the formulae.

Parallel Plate Capacitor

Two plates of area A at a small separation d: the field between them is \( E=\frac{\sigma}{\epsilon_0}=\frac Vd \) and \( C=\frac{\epsilon_0A}{d} \).
Force of attraction between the plates: \( F=\frac{q^2}{2\epsilon_0A}=\frac12QE \), where E is the field due to the full capacitor.
Battery connected: V is fixed and Q, U change. Battery disconnected: Q is fixed and V, U change. Many problems are solved just by deciding which of the two is held constant.
Work done in changing the separation equals the change in stored energy when the plates are moved slowly in an isolated capacitor; if the battery stays connected, the battery also does work.

Combination of Capacitors

Series: the same charge on every capacitor and the voltages add, \( \frac1{C_s}=\frac1{C_1}+\frac1{C_2}+\dots \); voltage divides in inverse proportion to capacitance.
Parallel: the same voltage across each and the charges add, \( C_p=C_1+C_2+\dots \).
Capacitors combine like conductances (the reverse of resistors). For irregular networks use symmetry or write charge conservation at every isolated node. Unlike plates joined together add charge; opposite plates joined subtract it.

RC Circuits

The charge on a capacitor cannot change suddenly if there is a resistor in series. An uncharged capacitor acts like a wire at t = 0 and like an open circuit (zero current) in the steady state.
Charging through R from a battery E: \( q=CE(1-e^{-t/\tau}),\ i=\frac ERe^{-t/\tau} \) with time constant \( \tau=RC \).
Discharging: \( q=q_0e^{-t/\tau},\ i=\frac{q_0}{RC}e^{-t/\tau} \). After one time constant the charge is 37% of its start. Half-life of the charge is \( \tau\ln2 \).
Energy balance while charging: the battery supplies \( CE^2 \) in total; half is stored and half is dissipated as heat in the resistor, regardless of R.

Capacitors with Dielectric and Partly Filled Gaps

A dielectric of constant K raises capacitance to \( C=\frac{K\epsilon_0A}{d} \) and reduces the field to \( E=\frac{\sigma}{K\epsilon_0} \). Induced surface charge: \( q_i=q\left(1-\frac1K\right) \).
Inserting a slab of thickness t (constant K) in a gap d: \( C=\frac{\epsilon_0A}{d-t+\frac tK} \). A conducting slab corresponds to K infinite, giving \( C=\frac{\epsilon_0A}{d-t} \).
Layers across the gap behave as capacitors in series, \( C=\frac{\epsilon_0A}{\sum d_i/K_i} \); layers side by side (partial areas) behave as capacitors in parallel, \( C=\frac{\epsilon_0}{d}\sum K_iA_i \).
Battery connected: inserting the slab keeps V, raises Q and U (U becomes KU). Battery disconnected: Q is fixed, V and U fall by the factor K, and the slab is pulled into the gap.

Other Types of Capacitors

Spherical capacitor (radii a < b, inner sphere charged): \( C=\frac{4\pi\epsilon_0ab}{b-a} \). If the outer shell is earthed this is the capacitance; if the shell is joined to the inner sphere the system behaves as one sphere \( 4\pi\epsilon_0b \).
Cylindrical capacitor of length l and radii a < b: \( C=\frac{2\pi\epsilon_0l}{\ln(b/a)} \).
A stack of n parallel plates connected alternately gives n − 1 capacitors in parallel, \( C=\frac{(n-1)\epsilon_0A}{d} \).

Worked Examples

Worked Example 1. A 5 \( \mu \)F conductor carries 30 \( \mu \)C and is joined by a wire to a distant isolated 10 \( \mu \)F conductor carrying 50 \( \mu \)C. Find the common potential, the heat produced and the final charges.
Answer:
Common potential: \( V=\frac{30+50}{5+10}=\frac{16}{3}\ \text{V} \).
Initial potentials 6 V and 5 V, so \( H=\frac12\cdot\frac{5\times10}{15}(1)^2=\frac53\ \mu\text{J} \).
Final charges: \( \frac{80}{3} \) \( \mu \)C on the 5 \( \mu \)F and \( \frac{160}{3} \) \( \mu \)C on the 10 \( \mu \)F conductor.

Worked Example 2. A 4 \( \mu \)F capacitor is charged to 100 V and the battery removed. A slab of constant 5 is then slipped in to fill the gap. Find the new voltage and the change in energy.
Answer:
Charge \( Q=400\ \mu\text{C} \) stays fixed. New capacitance 20 \( \mu \)F, so \( V'=\frac{400}{20}=20\ \text{V} \).
Initial energy \( \frac12(4)(100)^2=20\ \text{mJ} \); final \( \frac12(20)(20)^2=4\ \text{mJ} \). The energy falls by 16 mJ (work done by the field on the slab).

Worked Example 3. A 2 \( \mu \)F capacitor is charged through a 500 \( \Omega \) resistor from a 10 V battery. How long does it take to reach 6.3 V (take \( 1-e^{-1}=0.63 \))?
Answer:
\( \tau=RC=500\times2\times10^{-6}=1\ \text{ms} \).
6.3 V is 63% of 10 V, which is reached at \( t=\tau=1\ \text{ms} \).